a) x³ - 64x = 0

x(x² - 64) = 0

x(x - 8)(x + 8) = 0

x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0

*) x - 8 = 0

x = 8

*) x + 8 = 0

x = -8

Vậy x = -8; x = 0; x = 8

b) x³ - 4x² = -4x

x³ - 4x² + 4x = 0

x(x² - 4x + 4) = 0

x(x - 2)² = 0

x = 0 hoặc (x - 2)² = 0

*) (x - 2)² = 0

x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) x² - 16 - (x - 4) = 0

(x - 4)(x + 4) - (x - 4) = 0

(x - 4)(x + 4 - 1) = 0

(x - 4)(x + 3) = 0

x - 4 = 0 hoặc x + 3 = 0

*) x - 4 = 0

x = 4

*) x + 3 = 0

x = -3

Vậy x = -3; x = 4

d) (2x + 1)² = (3 + x)²

(2x + 1)² - (3 + x)² = 0

(2x + 1 - 3 - x)(2x + 1 + 3 + x) = 0

(x - 2)(3x + 4) = 0

x - 2 = 0 hoặc 3x + 4 = 0

*) x - 2 = 0

x = 2

*) 3x + 4 = 0

3x = -4

x = -4/3

Vậy x = -4/3; x = 2

e) x³ - 6x² + 12x - 8 = 0

(x - 2)³ = 0

x - 2 = 0

x = 2

f) x³ - 7x - 6 = 0

x³ + 2x² - 2x² - 4x - 3x - 6 = 0

(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0

x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0

(x + 2)(x² - 2x - 3) = 0

(x + 2)(x² + x - 3x - 3) = 0

(x + 2)[(x² + x) - (3x + 3)] = 0

(x + 2)[x(x + 1) - 3(x + 1)] = 0

(x + 2)(x + 1)(x - 3) = 0

x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x + 1 = 0

x = -1

*) x - 3 = 0

x = 3

Vậy x = -1; x = -1; x = 3

a,x\(^3\)-64=0
   x\(^3\)     =64
  =>x=3
b,x\(^3\)-4x\(^2\)=-4x

   x\(^3\)-4x\(^2\)+4x=0
   x(x\(^2\)-4x+4)=0
   x(x-2)\(^2\)=)
TH1:x=0
TH2:x-2=0
     =>x=2

c,x\(^2\)-16-(x-4)=0

   (x+4)(x-4)-(x-4)=0
   (x-4)(x+4-1)=0

   (x-4)(x+3)=0
TH1:x-4=0
     =>x=4

TH2:x+3=0

    =>x=-3

d,(2x+1).2=3+x
    4x+2-3-x=0
    3x-1=0
     x=\(\dfrac{1}{3}\)

e,x\(^3\)-6x\(^2\)+12x-8=0
   (x-2)\(^3\)=0
=>x-2=0
=>x=2

f,x\(^3\)-7x+6=0
  x\(^3\)-x-6x+6=0
  x(x\(^2\)-1)-6(x-1)=0
  x(x+1)(x-1)-6(x-1)=0
  (x-1)(x\(^2\)+x-6)=0
TH1:x-1=0

  =>x=1
TH2:x\(^2\)+x-6=0

       x\(^2\)+3x-2x-6=0

       x(x+3)-2(x+3)=0

       (x+3)(x-2)=0

=>x+3=0                =>x-2=0

+>x=-3                   =>x=2

d,(2x+1)\(^2\)=(3+x)\(^2\)

4x\(^2\)+4x+1-9-6x-x\(^2\)=0
3x\(^2\)-2x-8=0
3x\(^2\)-6x+4x-8=0
3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

TH1:3x+4=0                     TH2:x-2=0
=>x=\(\dfrac{-4}{3}\)                              =>x=2

1,\(=4x\left(x-\dfrac{3}{2}\right)\)
2,\(=-7y^3\left[2x^2y\left(2y+x\right)+3\right]\)
3, = 4x(a-b)-6xy(a-b)
=2x(a-b)(2-3y)
4,
=3(2x+1)-(2x-5)(2x+1)
=(3-2x+5)(2x+1)
=(8-2x)(2x+1)
=2(4-x)(2x+1)
 

5: \(4\left(x-3\right)^2+2x\left(3-x\right)\)

\(=\left(x-3\right)\left(4x-12\right)-2x\left(x-3\right)\)

\(=\left(x-3\right)\left(2x-12\right)\)

\(=2\left(x-6\right)\left(x-3\right)\)

8: \(\left(3x-1\right)^2-\left(x+3\right)^2\)

\(=\left(3x-1-x-3\right)\left(3x-1+x+3\right)\)

\(=\left(2x-4\right)\left(4x+2\right)\)

\(=4\left(x-2\right)\left(2x+1\right)\)

Giải gấp nhé mấy bạn

đây nhé

bạn viết câu hỏi dưới dạng trực quan để mn dễ hiểu nhé!

a) \(\sqrt{x}=3\left(x\ge0\right)\Leftrightarrow x=9\)

b) \(\sqrt{x}=\sqrt{5}\left(x\ge0\right)\Leftrightarrow x=5\)

c) \(\sqrt{x}=0\left(x\ge0\right)\Leftrightarrow x=0\)

d) \(\sqrt{x}=-2\left(x\ge0\right)\Leftrightarrow x=\varnothing\)

e) \(\sqrt{x-2}=3\left(x\ge0\right)\Leftrightarrow x-2=9\Leftrightarrow x=11\)

g) \(\sqrt{2x-1}=5\left(x\ge0\right)\Leftrightarrow2x-1=25\Leftrightarrow2x=26\Leftrightarrow x=13\)

h) \(\sqrt{x-3}=0\left(x\ge0\right)\Leftrightarrow x-3=0\Leftrightarrow x=3\)

a: \(\sqrt{x}=3\)

nên x=9

b: \(\sqrt{x}=\sqrt{5}\)

nên x=5

c: \(\sqrt{x}=0\)

nên x=0

d: \(\sqrt{x}=-2\)

nên \(x\in\varnothing\)

e: \(\sqrt{x}-2=3\)

\(\Leftrightarrow\sqrt{x}=5\)

hay x=25

g: \(\sqrt{2x}-1=5\)

\(\Leftrightarrow2x=36\)

hay x=18

h: Ta có: \(\sqrt{x}-3=0\)

nên x=9

a. \(\sqrt{x}=3\)

<=> x = 3²

<=> x = 9

b. \(\sqrt{5}=\sqrt{5}\)

<=> 5 = 5

<=> x có vô số nghiệm

c. \(\sqrt{x}=0\)

<=> x = 0²

<=> x = 0

d. \(\sqrt{x}=-2\)

<=> x = (-2)²

<=> x = 4

e. TH₁: \(\sqrt{x}-2=3\)

<=> \(\sqrt{x}=3+2\)

<=> \(\sqrt{x}=5\)

<=> x = 5²

<=> x = 25

TH₂: \(\sqrt{x-2}=3\)

<=> x - 2 = 3²

<=> x - 2 = 9

<=> x = 9 + 2

<=> x = 11

g. TH₁: \(\sqrt{2x}-1=5\)

<=> \(\sqrt{2x}=5+1\)

<=> \(\sqrt{2x}=6\)

<=> 2x = 6²

<=> 2x = 36

<=> x = 18

TH₂: \(\sqrt{2x-1}=5\)

<=> 2x - 1 = 5²

<=> 2x - 1 = 25

<=> 2x = 25 + 1

<=> 2x = 26

<=> x = 13

h. TH₁: \(\sqrt{x}-3=0\)

<=> \(\sqrt{x}=0+3\)

<=> \(\sqrt{x}=3\)

<=> x = 3²

<=> x = 9

TH₂: \(\sqrt{x-3}=0\)

<=> x - 3 = 0²

<=> x - 3 = 0

<=> x = 0 + 3

<=> x = 3

(Lưu ý: các TH₁ và TH₂ là do mik không hiểu rõ đề, bn biết đề rồi thì chỉ cần làm theo phần đúng thôi nha.)

a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)

mà -3<x<30

nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)

b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)

mà -16<=x<20

nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)

c: \(\Leftrightarrow x-1+4⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

d: \(\Leftrightarrow2x+4-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

\(A=\sqrt{x^2-4x+25}=\sqrt{\left(x-2\right)^2+21}\)

Ta có :   \(\left(x-2\right)^2\ge0\) =>  \(\left(x-2\right)^2+21\ge21\left(\forall x\right)\) => \(\sqrt{\left(x-2\right)^2+21}\ge\sqrt{21}\left(\forall x\right)\)                 

Dấu " = "  xảy ra   \(\Leftrightarrow\)   \(\sqrt{\left(x-2\right)^2}=0\)            

                              \(\Leftrightarrow\)  \(x-2=0\)                  

                              \(\Leftrightarrow\)  x  =  2 

Vậy giá trị nhỏ nhất của A là :  \(\sqrt{21}\)      khi x  =  2

\(B=\sqrt{x^2-6x+30}=\sqrt{\left(x-3\right)^2+21}\)      

Vì   \(\sqrt{\left(x-3\right)^2}\ge0\left(\forall x\right)\)=> \(\sqrt{\left(x-3\right)^2+21}\ge\sqrt{21}\left(\forall x\right)\)                  

Dấu "  =  "  xảy ra  \(\Leftrightarrow\)   \(\sqrt{\left(x-3\right)^2}=0\)                          

                                \(\Leftrightarrow\)  \(x-3=0\)                      

                                \(\Leftrightarrow\) \(x=3\)                             

Vậy giá trị nhỏ nhất của B là :  \(\sqrt{21}\)  khi x  =  3

\(D=\sqrt{x^2-4x+7}+\sqrt{2}=\sqrt{\left(x-2\right)^2+3}+\sqrt{2}\)

Vì  

???????????????????????

\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)

a. (x + 2)² + (x + 3)² - 2(x - 2)(x - 3) = 19

<=> (x² + 4x + 4) + (x² + 6x + 9) - (2x + 4)(x - 3) = 19

<=> x² + 4x + 4 + x² + 6x + 9 - 2x² + 6x - 4x + 12 = 19

<=> x² + x² - 2x² + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0

<=> 12x + 6 = 0

<=> 6(2x + 1) = 0

<=> 2x + 1 = 0

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)  

b. (x + 2)(x² - 2x + 4) - x(x² - 5) = 15

<=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 5x - 15 = 0

<=> x³ - x³ - 2x² + 2x² + 4x - 4x + 5x + 8 - 15 = 0

<=> 5x - 7 = 0

<=> 5x = 7

<=> x = \(\dfrac{7}{5}\)